An object is attached to a hanging unstretched, ideal and massless spring and slowly lowered to its equilibrium position, a distance of 5.3 cm below the starting point. If instead of having been lowered slowly the object was dropped from rest, how far then would it then stretch the spring at maximum elongation (measured from the point it was dropped)?

Question

I got an answer of 12.8cm but this is incorrect, (textbook says 11cm rounded)

Answer 1

lower ... m g = k x ... weight = spring force

drop ... m g y = 1/2 k y^2 ... potential energy at top = work to stretch spring

2 m g = k y ... substituting ... 2 k x = k y ... 2 x = y

Answer 2

Hey, Thanks so much for answering!!